Question: If $x \boxdot y = xy+4x-y$ and $x \diamond y = x^{2}+4y^{2}$, find $(1 \boxdot -3) \diamond -1$.
Answer: First, find $1 \boxdot -3$ $ 1 \boxdot -3 = -3+(4)(1)-(-3)$ $ \hphantom{1 \boxdot -3} = 4$ Now, find $4 \diamond -1$ $ 4 \diamond -1 = 4^{2}+4(-1)^{2}$ $ \hphantom{4 \diamond -1} = 20$.